使用Spring时如何注入多个JPA EntityManager(持久化单元)

问题描述

我需要使用一个数据库进行查询(非修改)和一个用于命令(修改).我使用的是 Spring Data JPA，所以我有两个配置类:

I need to use one database for queries (non-modifying) and one for commands (modifying). I am using Spring Data JPA, so I have two configuration classes:

@Configuration
@EnableJpaRepositories(value = "com.company.read",
        entityManagerFactoryRef = "readingEntityManagerFactory",
        transactionManagerRef = "readingTransactionManager")
@EnableTransactionManagement
public class SpringDataJpaReadingConfiguration {

    @Bean(name = "readingEntityManagerFactory")
    public EntityManagerFactory readingEntityManagerFactory() {
        return Persistence.createEntityManagerFactory("persistence.reading");
    }

    @Bean(name = "readingExceptionTranslator")
    public HibernateExceptionTranslator readingHibernateExceptionTranslator() {
        return new HibernateExceptionTranslator();
    }

    @Bean(name = "readingTransactionManager")
    public JpaTransactionManager readingTransactionManager() {
        return new JpaTransactionManager();
    }

}

@Configuration
@EnableJpaRepositories(value = "com.company.write",
        entityManagerFactoryRef = "writingEntityManagerFactory",
        transactionManagerRef = "writingTransactionManager")
@EnableTransactionManagement
public class SpringDataJpaWritingConfiguration {

    @Bean(name = "writingEntityManagerFactory")
    public EntityManagerFactory writingEntityManagerFactory() {
        return Persistence.createEntityManagerFactory("persistence.writing");
    }

    @Bean(name = "writingExceptionTranslator")
    public HibernateExceptionTranslator writingHibernateExceptionTranslator() {
        return new HibernateExceptionTranslator();
    }

    @Bean(name = "writingTransactionManager")
    public JpaTransactionManager writingTransactionManager() {
        return new JpaTransactionManager();
    }

}

在我的存储库中，我有时需要使用 EntityManager 来决定这样使用:

In my repository I sometimes need to decide with EntityManager to use like so:

@Repository
public class UserReadingRepository {

    @PersistenceContext(unitName = "persistence.reading")
    private EntityManager em;

    // some useful queries here
}

我正在使用 persistence.xml 中定义的持久性单元名称:

I am using persistence unit's name as defined in my persistence.xml:

<persistence xmlns="http://www.51sjk.com/Upload/Articles/1/0/336/336201_20221104102808798.jpg"
             xmlns:xsi="http://www.51sjk.com/Upload/Articles/1/0/336/336201_20221104102809579.jpg"
             xsi:schemaLocation="http://www.51sjk.com/Upload/Articles/1/0/336/336201_20221104102808798.jpg http://www.51sjk.com/Upload/Articles/1/0/336/336201_20221104102808798.jpg/persistence_2_0.xsd"
             version="2.0">

    <persistence-unit name="persistence.reading" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <non-jta-data-source>ReadingDS</non-jta-data-source>
        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
            <property name="hibernate.show_sql" value="true" />
        </properties>
    </persistence-unit>

    <persistence-unit name="persistence.writing" transaction-type="RESOURCE_LOCAL">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
        <non-jta-data-source>WritingDS</non-jta-data-source>
        <properties>
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
            <property name="hibernate.show_sql" value="true" />
        </properties>
    </persistence-unit>

</persistence>

Spring 抛出 org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' 被定义.奇怪的是，Spring 似乎试图用持久性单元名称实例化 a bean?我是不是配置错了什么?

Spring throws org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined. Oddly, it looks like Spring tries to instantiate a bean with persistence unit name? Did I misconfigure something?

UPDATE:当我从 @PersistenceContext 注释中删除 unitName = "persistence.reading" 时，我会收到以下错误:org.springframework.beans.factory.NoUniqueBeanDefinitionException:没有定义 [javax.persistence.EntityManagerFactory] ??类型的合格 bean:预期的单个匹配 bean 但找到了 2:readingEntityManagerFactory,writingEntityManagerFactory

UPDATE: When I remove unitName = "persistence.reading" from @PersistenceContext annotation, I will get following error instead: org.springframework.beans.factory.NoUniqueBeanDefinitionException: No qualifying bean of type [javax.persistence.EntityManagerFactory] is defined: expected single matching bean but found 2: readingEntityManagerFactory,writingEntityManagerFactory

更新 2:Rohit 建议(在评论中)改为连接 EntityManagerFactory.所以我尝试执行以下操作:

UPDATE 2: Rohit suggested (in the comment) to wire EntityManagerFactory instead. So I tried to do the following:

@PersistenceUnit(unitName = "persistence.reading")
private EntityManagerFactory emf;

但 Spring 只报告:org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined

but Spring only reports: org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'persistence.reading' is defined

最终修复:感谢 Vlad 的回答，我能够更新代码以使用以下内容(只需确保您也定义了 dataSource bean):

FINAL FIX: Thanks to Vlad's answer, I was able to update the code to use the following (just make sure you define your dataSource bean as well):

@Bean(name = "readingEntityManagerFactory")
public EntityManagerFactory readingEntityManagerFactory() {
    LocalContainerEntityManagerFactoryBean em = new LocalContainerEntityManagerFactoryBean();
    em.setPersistenceUnitName("persistence.reading");
    em.setDataSource(dataSource());
    em.setPackagesToScan("com.company");
    em.setJpaVendorAdapter(new HibernateJpaVendorAdapter());
    em.afterPropertiesSet();
    return em.getObject();
}

问题描述

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